Question: You have found the following ages (in years) of all $4$ gorillas at your local zoo: $ 7,\enspace 10,\enspace 15,\enspace 21$ What is the average age of the gorillas at your zoo? What is the variance? Round your answers to the nearest tenth. Average age: $ $
Answer: Because we have data for all $4$ gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$. To find the population mean, add up the values of all $4$ ages and divide by $4$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 10 + 15 + 21}{{4}} = {13.25\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-6.25$ years $39.06$ years $^2$ $10$ years $-3.25$ years $10.56$ years $^2$ $15$ years $1.75$ years $3.06$ years $^2$ $21$ years $7.75$ years $60.06$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the population variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{39.06} + {10.56} + {3.06} + {60.06}} {{4}} $ $ {\sigma^2} = \dfrac{{112.74}}{{4}} = {28.19\text{ years}^2} $ The average gorilla at the zoo is $13.3$ years old. The population variance is $28.2$ years $^2$.